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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times. Example 1: Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. Example 2: Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10]. /** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ public class Solution { public ArrayList insert(ArrayList intervals, Interval newInterval) { Interval temp = newInterval; ArrayList lst = new ArrayList(); for(Interval i : intervals) { if(i.end < temp.start) { lst.add(i); } else if(i.start > temp.end){ lst.add(temp); temp = i; } else { temp = new Interval(Math.min(temp.start, i.start), Math.max(temp.end, i.end)); } } lst.add(temp); return lst; } }